CS61A_lab03

题目描述：

代码：

 1 def num_eights(x):
 2     """Returns the number of times 8 appears as a digit of x.
 3 
 4     >>> num_eights(3)
 5     0
 6     >>> num_eights(8)
 7     1
 8     >>> num_eights(88888888)
 9     8
10     >>> num_eights(2638)
11     1
12     >>> num_eights(86380)
13     2
14     >>> num_eights(12345)
15     0
16     >>> from construct_check import check
17     >>> # ban all assignment statements
18     >>> check(HW_SOURCE_FILE, 'num_eights',
19     ...       ['Assign', 'AugAssign'])
20     True
21     """
22     "*** YOUR CODE HERE ***"
23     if x==0:
24         return 0
25     return num_eights(x//10)+(x%10==8)
26     
27 
28 
29 def pingpong(n):
30     """Return the nth element of the ping-pong sequence.
31 
32     >>> pingpong(8)
33     8
34     >>> pingpong(10)
35     6
36     >>> pingpong(15)
37     1
38     >>> pingpong(21)
39     -1
40     >>> pingpong(22)
41     -2
42     >>> pingpong(30)
43     -2
44     >>> pingpong(68)
45     0
46     >>> pingpong(69)
47     -1
48     >>> pingpong(80)
49     0
50     >>> pingpong(81)
51     1
52     >>> pingpong(82)
53     0
54     >>> pingpong(100)
55     -6
56     >>> from construct_check import check
57     >>> # ban assignment statements
58     >>> check(HW_SOURCE_FILE, 'pingpong', ['Assign', 'AugAssign'])
59     True
60     """
61     "*** YOUR CODE HERE ***"
62     #special case :
63     #1. n%8==0 2. num_eights(n)>0
64     #我的问题在于如何确定是正在递增还是正在递减呢
65     #肯定是返回前一个数的函数再加上1或者减去1
66     #如果是正在递增，则返回上一个数的函数+1，如果是递减，则返回上一个数的函数-1
67     def helper(i,value,direction):
68         if i==n:
69             return value 
70         if num_eights(i)!=0 or i%8==0:
71             return helper(i+1,value-direction,direction*-1)
72         #用来改变方向
73         else :
74             return helper(i+1,value+direction,direction)
75     return helper(1,1,1)
76     #从起点开始往后递归，而不是从终点往前递归
77     #启示：如果只有一个参数不够反应情况，那么就可以再重新定义一个函数

启示：

  • 递归不一定是要从后往前递归，也有可能是从前往后递归
  • If you need to keep track of more than one value across recursive calls, consider writing a helper function.
posted @ 2023-02-15 20:17  哎呦_不想学习哟~  阅读(22)  评论(0)    收藏  举报